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  •  A particle of mass 1 kg is placed at a distance of m from centre  on an axis of a uniform dis

 A particle of mass 1 kg is placed at a distance of 3\sqrt{3} m from centre  on an axis of a uniform disc of mass 5 kg and radius 3 m. If gravitational potential due to uniform disc at this position is V. If we decrease the distance of particle to 4m from its centre along its axis . Now  what is the gravitational potential  due to uniform disc at this new position is ( in V ) (give answer till 1 decimal place)

Option: 1

1.3


Option: 2

0.5


Option: 3

2.6


Option: 4

11


Answers (1)

best_answer

As we learned 

Gravitational Potential due to Uniform disc -

For Uniform disc

 

 

a \rightarrow Radius of disc

M-mass of disc

  • At the centre of the disc 

              V=-\frac{2GM}{a}

  • At a point on its axis

                    V=-\frac{2GM}{a^2}(\sqrt{a^2+x^2}-x)

So

So putting x= 3\sqrt{3}m and a=3m ,M=1 kg

We get V_B=V=-\frac{2G*1}{3^2}(\sqrt{3^2+(3\sqrt{3})^2}-3\sqrt{3})=-\frac{2G}{9}*(0.803)=-0.17G

Now putting x=4m 

V_ A=-\frac{2G*1}{3^2}(\sqrt{3^2+(4)^2}-4)=-\frac{2G}{9}*(1)=-0.22G

Taking ratio we get 

\frac{V}{V_A}=\frac{0.17}{0.22}\\ \Rightarrow V_A=1.29 V\approx 1.3 V

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